The electric potential at any point as a function of distance $(x)$ in meters is given by $V = 5x^2 + 10x - 9 \, (V)$. The value of the electric field at $x = 1 \, m$ is ...... $V m^{-1}$.

The electric field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $\theta = 60^o$ with respect to the field lines as shown in the figure. The potential difference between points $1$ and $2$ is

The electric potential $V$ at any point $(x, y, z)$ (all in $m$) in space is given by $V = 4x^2 \ V$. The electric field at the point $(1 \ m, 0, 2 \ m)$ in $V/m$ is:

An infinite non-conducting sheet has a surface charge density of $7 \times 10^{-7} \text{ C m}^{-2}$ on one side. The distance between equipotential surfaces whose potentials differ by $19.8 \text{ V}$ will be (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^9 \text{ SI units}$) (in $\text{ mm}$)

If the electric potential is given by $V = (5x^2 + 10x - 9) \ V$,what is the electric field at $x = 1 \ m$ in $V/m$?

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?